Assignment 12
Table of contents
url: https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/resources/mit18_100af20_hw12/
Chapters and exercises given with a numbering are from Basic Analysis: Introduction to Real Analysis (Vol I) by J. Lebl.
Exercise 6.1.2
a) Find the pointwise limit $\frac{e^{x/n}}{n} $ for $x\in \mathbb{R}$
$ f_n(x) := \frac{e^{x/n}}{n}, \quad f(x) := \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{e^{x/n}}{n} = 0, \ \forall x \in \mathbb{R} $
b) Is the limit uniform on $\mathbb{R}$?
Let $ \varepsilon = \frac{1}{4} $, $ x_n := n \log n $. Then $f_n(x_n) = \frac{e^{x_n/n}}{n} = \frac{e^{\log n}}{n} = 1$,
\[\vert f_n(x_n) - f(x)\vert = 1 > \varepsilon = \frac{1}{4}\]$\therefore$ Limit is not uniform on $ \mathbb{R} $.
c) Is the limit uniform on $[0,1]$?
$ x \in [0,1] \Rightarrow \frac{x}{n} \in [0, \frac{1}{n}] \Rightarrow e^{x/n} \leq e^{1/n} $
Let $ M = \left\lceil \frac{1}{\log \varepsilon} \right\rceil \in \mathbb{N} $, then $ \forall n \geq M $,
\[\vert f_n(x) - f(x)\vert = \left\vert \frac{e^{x/n}}{n} \right\vert \leq \frac{e^{1/n}}{n} \leq \frac{e^{1/M}}{M} < \varepsilon\]$\therefore$ Limit is uniform on $ [0,1] $.
Exercise 6.1.5
Suppose \(\{f_n\}_{n=1}^\infty\) and \(\{g_n\}_{n=1}^\infty\) converge uniformly to $f$ and $g$ on $A$, respectively. Show that \(\{f_n + g_n\}_{n=1}^\infty\) converges uniformly to $f + g$ on $A$.
Let $\varepsilon > 0$. Since $f_n \to f$ uniformly, $\exists N_1 \in \mathbb{N}$, such that $\forall n \geq N_1, \forall x \in A,$. \(\vert f_n(x) - f(x)\vert < \varepsilon/2\) Similarly, since $g_n \to g$ uniformly, $\exists N_2 \in \mathbb{N}$, such that $ \forall n \geq N_2, \forall x \in A$, \(\vert g_n(x) - g(x)\vert < \varepsilon/2\)
Let $N = \max(N_1, N_2)$. Then for all $n \geq N$, and all $x \in A$, \(\vert (f_n + g_n)(x) - (f + g)(x)\vert \leq \vert f_n(x) - f(x)\vert + \vert g_n(x) - g(x)\vert < \varepsilon\) Hence, \(\lim_{n \to \infty} \sup_{x \in A} \vert (f_n + g_n)(x) - (f + g)(x)\vert = 0\)